Sachpazis pad footing example

GEODOMISI Ltd. - Dr. Costas Sachpazis
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
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Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Project
Pad footing analysis and design (BS8110-1:1997)
Job Ref.
Section
Civil & Geotechnical Engineering
Sheet no./rev.
1
Calc. by
Dr.C.Sach
pazis
Date
23/05/2013
Chk'd by
-
Date App'd by Date
PAD FOOTING ANALYSIS AND DESIGN (BS8110-1:1997)
Pad footing details
Length of pad footing; L = 2500 mm
Width of pad footing; B = 1500 mm
Area of pad footing; A = L × B = 3.750 m
2
Depth of pad footing; h = 400 mm
Depth of soil over pad footing; hsoil = 200 mm
Density of concrete; ρconc = 23.6 kN/m
3
Column details
Column base length; lA = 300 mm
Column base width; bA = 300 mm
Column eccentricity in x; ePxA = 0 mm
Column eccentricity in y; ePyA = 0 mm
Soil details
Dense, moderately graded, sub-angular, gravel
Mobilisation factor; m= ;1.5;
Density of soil; ρsoil = 20.0 kN/m
3
Design shear strength; φ’ = 25.0 deg
Design base friction; δ = 19.3 deg
Allowable bearing pressure; Pbearing = 200 kN/m
2
Axial loading on column
Dead axial load on column; PGA = 200.0 kN
2500
1100 1100

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Date
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Imposed axial load on column; PQA = 165.0 kN
Wind axial load on column; PWA = 0.0 kN
Total axial load on column; PA = 365.0 kN
Foundation loads
Dead surcharge load; FGsur = 0.000 kN/m
2
Imposed surcharge load; FQsur = 0.000 kN/m
2
Pad footing self weight; Fswt = h × ρconc = 9.440 kN/m
2
Soil self weight; Fsoil = hsoil × ρsoil = 4.000 kN/m
2
Total foundation load; F = A × (FGsur + FQsur + Fswt + Fsoil) = 50.4 kN
Horizontal loading on column base
Dead horizontal load in x direction; HGxA = 20.0 kN
Imposed horizontal load in x direction; HQxA = 15.0 kN
Wind horizontal load in x direction; HWxA = 0.0 kN
Total horizontal load in x direction; HxA = 35.0 kN
Dead horizontal load in y direction; HGyA = 5.0 kN
Imposed horizontal load in y direction; HQyA = 5.0 kN
Wind horizontal load in y direction; HWyA = 0.0 kN
Total horizontal load in y direction; HyA = 10.0 kN
Moment on column base
Dead moment on column in x direction; MGxA = 15.000 kNm
Imposed moment on column in x direction; MQxA = 10.000 kNm
Wind moment on column in x direction; MWxA = 0.000 kNm
Total moment on column in x direction; MxA = 25.000 kNm
Dead moment on column in y direction; MGyA = 25.000 kNm
Imposed moment on column in y direction; MQyA = 30.000 kNm
Wind moment on column in y direction; MWyA = 0.000 kNm
Total moment on column in y direction; MyA = 55.000 kNm
Check stability against sliding
Resistance to sliding due to base friction
Hfriction = max([PGA + (FGsur + Fswt + Fsoil) × A], 0 kN) × tan(δ) = 87.7 kN
Passive pressure coefficient; Kp = (1 + sin(φ’)) / (1 - sin(φ’)) = 2.464
Stability against sliding in x direction
Passive resistance of soil in x direction; Hxpas = 0.5 × Kp × (h
2
+ 2 × h × hsoil) × B × ρsoil =
11.8 kN
Total resistance to sliding in x direction; Hxres = Hfriction + Hxpas = 99.5 kN
PASS - Resistance to sliding is greater than horizontal load in x direction
Stability against sliding in y direction
Passive resistance of soil in y direction; Hypas = 0.5 × Kp × (h
2
+ 2 × h × hsoil) × L × ρsoil =
19.7 kN
Total resistance to sliding in y direction; Hyres = Hfriction + Hypas = 107.4 kN

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Dr.C.Sach
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Date
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PASS - Resistance to sliding is greater than horizontal load in y direction
Check stability against overturning in x direction
Total overturning moment; MxOT = MxA + HxA × h = 39.000 kNm
Restoring moment in x direction
Foundation loading; Mxsur = A × (FGsur + Fswt + Fsoil) × L / 2 = 63.000
kNm
Axial loading on column; Mxaxial = (PGA) × (L / 2 - ePxA) = 250.000 kNm
Total restoring moment; Mxres = Mxsur + Mxaxial = 313.000 kNm
PASS - Restoring moment is greater than overturning moment in x direction
Check stability against overturning in y direction
Total overturning moment; MyOT = MyA + HyA × h = 59.000 kNm
Restoring moment in y direction
Foundation loading; Mysur = A × (FGsur + Fswt + Fsoil) × B / 2 = 37.800
kNm
Axial loading on column; Myaxial = (PGA) × (B / 2 - ePyA) = 150.000 kNm
Total restoring moment; Myres = Mysur + Myaxial = 187.800 kNm
PASS - Restoring moment is greater than overturning moment in y direction
Calculate pad base reaction
Total base reaction; T = F + PA = 415.4 kN
Eccentricity of base reaction in x; eTx = (PA × ePxA + MxA + HxA × h) / T = 94 mm
Eccentricity of base reaction in y; eTy = (PA × ePyA + MyA + HyA × h) / T = 142 mm
Check pad base reaction eccentricity
abs(eTx) / L + abs(eTy) / B = 0.132
Base reaction acts within combined middle third of base
Calculate pad base pressures
q1 = T / A - 6 × T × eTx / (L × A) - 6 × T × eTy / (B ×
A) = 22.880 kN/m
2
q2 = T / A - 6 × T × eTx / (L × A) + 6 × T × eTy / (B ×
A) = 148.747 kN/m
2
q3 = T / A + 6 × T × eTx / (L × A) - 6 × T × eTy / (B ×
A) = 72.800 kN/m
2
q4 = T / A + 6 × T × eTx / (L × A) + 6 × T × eTy / (B ×
A) = 198.667 kN/m
2
Minimum base pressure; qmin = min(q1, q2, q3, q4) = 22.880 kN/m
2
Maximum base pressure; qmax = max(q1, q2, q3, q4) = 198.667 kN/m
2
PASS - Maximum base pressure is less than allowable bearing pressure

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Partial safety factors for loads
Partial safety factor for dead loads; γfG = 1.40
Partial safety factor for imposed loads; γfQ = 1.60
Partial safety factor for wind loads; γfW = 0.00
Ultimate axial loading on column
Ultimate axial load on column; PuA = PGA × γfG + PQA × γfQ + PWA × γfW = 544.0 kN
Ultimate foundation loads
Ultimate foundation load; Fu = A × [(FGsur + Fswt + Fsoil) × γfG + FQsur × γfQ] =
70.6 kN
Ultimate horizontal loading on column
Ultimate horizontal load in x direction; HxuA = HGxA × γfG + HQxA × γfQ + HWxA × γfW = 52.0
kN
Ultimate horizontal load in y direction; HyuA = HGyA × γfG + HQyA × γfQ + HWyA × γfW = 15.0
kN
Ultimate moment on column
Ultimate moment on column in x direction; MxuA = MGxA × γfG + MQxA × γfQ + MWxA × γfW =
37.000 kNm
Ultimate moment on column in y direction; MyuA = MGyA × γfG + MQyA × γfQ + MWyA × γfW =
83.000 kNm
22.9 kN/m
148.7 kN/m
72.8 kN/m
198.7 kN/m
2
2
2
2

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Sheet no./rev.
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Calc. by
Dr.C.Sach
pazis
Date
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Chk'd by
-
Date App'd by Date
Calculate ultimate pad base reaction
Ultimate base reaction; Tu = Fu + PuA = 614.6 kN
Eccentricity of ultimate base reaction in x; eTxu = (PuA × ePxA + MxuA + HxuA × h) / Tu = 94 mm
Eccentricity of ultimate base reaction in y; eTyu = (PuA × ePyA + MyuA + HyuA × h) / Tu = 145 mm
Calculate ultimate pad base pressures
q1u = Tu/A - 6×Tu×eTxu/(L×A) - 6×Tu×eTyu/(B×A) =
31.957 kN/m
2
q2u = Tu/A - 6×Tu×eTxu/(L×A) + 6×Tu× eTyu/(B×A) =
221.824 kN/m
2
q3u = Tu/A + 6×Tu×eTxu/(L×A) - 6×Tu×eTyu/(B×A) =
105.941 kN/m
2
q4u = Tu/A + 6×Tu×eTxu/(L×A) + 6×Tu×eTyu/(B×A) =
295.808 kN/m
2
Minimum ultimate base pressure; qminu = min(q1u, q2u, q3u, q4u) = 31.957 kN/m
2
Maximum ultimate base pressure; qmaxu = max(q1u, q2u, q3u, q4u) = 295.808 kN/m
2
Calculate rate of change of base pressure in x direction
Left hand base reaction; fuL = (q1u + q2u) × B / 2 = 190.336 kN/m
Right hand base reaction; fuR = (q3u + q4u) × B / 2 = 301.312 kN/m
Length of base reaction; Lx = L = 2500 mm
Rate of change of base pressure; Cx = (fuR - fuL) / Lx = 44.390 kN/m/m
Calculate pad lengths in x direction
Left hand length; LL = L / 2 + ePxA = 1250 mm
Right hand length; LR = L / 2 - ePxA = 1250 mm
Calculate ultimate moments in x direction
Ultimate moment in x direction; Mx = fuL×LL
2
/2+Cx×LL
3
/6-Fu×LL
2
/(2×L)+HxuA×h+MxuA
= 198.900 kNm
Calculate rate of change of base pressure in y direction
Top edge base reaction; fuT = (q2u + q4u) × L / 2 = 647.040 kN/m
Bottom edge base reaction; fuB = (q1u + q3u) × L / 2 = 172.373 kN/m
Length of base reaction; Ly = B = 1500 mm
Rate of change of base pressure; Cy = (fuB - fuT) / Ly = -316.444 kN/m/m
Calculate pad lengths in y direction
Top length; LT = B / 2 - ePyA = 750 mm
Bottom length; LB = B / 2 + ePyA = 750 mm
Calculate ultimate moments in y direction
Ultimate moment in y direction; My = fuT×LT
2
/2+Cy×LT
3
/6-Fu×LT
2
/(2×B) = 146.500
kNm
Material details
Characteristic strength of concrete; fcu = 30 N/mm
2

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Characteristic strength of reinforcement; fy = 500 N/mm
2
Characteristic strength of shear reinforcement; fyv = 500 N/mm
2
Nominal cover to reinforcement; cnom = 30 mm
Moment design in x direction
Diameter of tension reinforcement; φxB = 12 mm
Depth of tension reinforcement; dx = h - cnom - φxB / 2 = 364 mm
Design formula for rectangular beams (cl 3.4.4.4)
Kx = Mx / (B × dx
2
× fcu) = 0.033
Kx’ = 0.156
Kx < Kx' compression reinforcement is not required
Lever arm; zx = dx × min([0.5 + √(0.25 - Kx / 0.9)], 0.95) = 346
mm
Area of tension reinforcement required; As_x_req = Mx / (0.87 × fy × zx) = 1322 mm
2
Minimum area of tension reinforcement; As_x_min = 0.0013 × B × h = 780 mm
2
Tension reinforcement provided; 12 No. 12 dia. bars bottom (125 centres)
Area of tension reinforcement provided; As_xB_prov = NxB × π × φxB
2
/ 4 = 1357 mm
2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Moment design in y direction
Diameter of tension reinforcement; φyB = 12 mm
Depth of tension reinforcement; dy = h - cnom - φxB - φyB / 2 = 352 mm
Design formula for rectangular beams (cl 3.4.4.4)
Ky = My / (L × dy
2
× fcu) = 0.016
Ky’ = 0.156
Ky < Ky' compression reinforcement is not required
Lever arm; zy = dy × min([0.5 + √(0.25 - Ky / 0.9)], 0.95) = 334
mm
Area of tension reinforcement required; As_y_req = My / (0.87 × fy × zy) = 1007 mm
2
Minimum area of tension reinforcement; As_y_min = 0.0013 × L × h = 1300 mm
2
Tension reinforcement provided; 13 No. 12 dia. bars bottom (200 centres)
Area of tension reinforcement provided; As_yB_prov = NyB × π × φyB
2
/ 4 = 1470 mm
2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Calculate ultimate shear force at d from right face of column
Ultimate pressure for shear; qsu = (q1u + Cx × (L / 2 + ePxA + lA / 2 + dx) / B + q4u)
/ 2
qsu = 189.984 kN/m
2
Area loaded for shear; As = B × min(3 × (L / 2 - eTx), L / 2 - ePxA - lA / 2 - dx)
= 1.104 m
2
Ultimate shear force; Vsu = As × (qsu - Fu / A) = 188.970 kN
Shear stresses at d from right face of column (cl 3.5.5.2)
Design shear stress; vsu = Vsu / (B × dx) = 0.346 N/mm
2

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Sheet no./rev.
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Calc. by
Dr.C.Sach
pazis
Date
23/05/2013
Chk'd by
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Date App'd by Date
From BS 8110:Part 1:1997 - Table 3.8
Design concrete shear stress; vc = 0.432 N/mm
2
Allowable design shear stress; vmax = min(0.8N/mm
2
× √(fcu / 1 N/mm
2
), 5 N/mm
2
)
= 4.382 N/mm
2
PASS - vsu < vc - No shear reinforcement required
Calculate ultimate punching shear force at face of column
Ultimate pressure for punching shear; qpuA = q1u+[(L/2+ePxA-lA/2)+(lA)/2]×Cx/B-[(B/2+ePyA-
bA/2)+(bA)/2]×Cy/L
qpuA = 163.883 kN/m
2
Average effective depth of reinforcement; d = (dx + dy) / 2 = 358 mm
Area loaded for punching shear at column; ApA = (lA)×(bA) = 0.090 m
2
Length of punching shear perimeter; upA = 2×(lA)+2×(bA) = 1200 mm
Ultimate shear force at shear perimeter; VpuA = PuA + (Fu / A - qpuA) × ApA = 530.944 kN
Effective shear force at shear perimeter; VpuAeff =
VpuA×[1+1.5×abs(MxuA)/(VpuA×(bA))+1.5×abs(MyuA)/(VpuA×(lA))] = 1130.944 kN
Punching shear stresses at face of column (cl 3.7.7.2)
Design shear stress; vpuA = VpuAeff / (upA × d) = 2.633 N/mm
2
2
× √(fcu / 1 N/mm
2
), 5 N/mm
2
)
= 4.382 N/mm
2
PASS - Design shear stress is less than allowable design shear stress
Calculate ultimate punching shear force at perimeter of 1.5 d from face of column
Ultimate pressure for punching shear; qpuA1.5d = q1u+[(L/2+ePxA-lA/2-
1.5×d)+(lA+2×1.5×d)/2]×Cx/B-[B/2]×Cy/L
qpuA1.5d = 163.883 kN/m
2
Average effective depth of reinforcement; d = (dx + dy) / 2 = 358 mm
Area loaded for punching shear at column; ApA1.5d = (lA+2×1.5×d)×B = 2.061 m
2
Length of punching shear perimeter; upA1.5d = 2×B = 3000 mm
Ultimate shear force at shear perimeter; VpuA1.5d = PuA + (Fu / A - qpuA1.5d) × ApA1.5d = 245.018 kN
Effective shear force at shear perimeter; VpuA1.5deff = VpuA1.5d × 1.25 = 306.272 kN
Punching shear stresses at perimeter of 1.5 d from face of column (cl 3.7.7.2)
Design shear stress; vpuA1.5d = VpuA1.5deff / (upA1.5d × d) = 0.285 N/mm
2
From BS 8110:Part 1:1997 - Table 3.8
Design concrete shear stress; vc = 0.409 N/mm
2
2
× √(fcu / 1 N/mm
2
), 5 N/mm
2
)
= 4.382 N/mm
2
PASS - vpuA1.5d < vc - No shear reinforcement required

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Sheet no./rev.
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Calc. by
Dr.C.Sach
pazis
Date
23/05/2013
Chk'd by
-
Date App'd by Date
Shear at d from column face
Punching shear perimeter at 1.5 × d from column face
13 No. 12 dia. bars btm (200 c/c)
12 No. 12 dia. bars btm (125 c/c)

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